Question

What will be the percentage decrease in weight of a body, When it is taken 32 km below the surface of earth.Take radius of earth 6400 km.

1 point


​

Answer 1

Answer :

• A body is taken 32 km above the surface of the earth
• Radius of earth = 6400 km
• Percentage decrease in weight of the body = ?

$\quad$━━━━━━━━━━━━━━━━━━

$\sf :\implies g' = g\bigg\lgroup 1 - \dfrac{h}{R}\bigg\rgroup\;\;-eq(1)$

• So then the % decrease in the value of acceleration due to gravity is gonna be,

$\sf :\implies \delta g = \dfrac{g - g'}{g} \times 100$

$\sf :\implies \delta g = \dfrac{g - g\bigg\lgroup 1 - \dfrac{h}{R}\bigg\rgroup}{g} \times 100$

$\sf :\implies \delta g = \dfrac{g\bigg\{ 1 - \bigg\lgroup 1 - \dfrac{h}{R}\bigg\rgroup\bigg\} }{g} \times 100$

$\sf :\implies \delta g = \bigg\{ 1 - 1 + \dfrac{h}{R} \bigg\} 100$

$\sf :\implies \delta g = \dfrac{100h}{R} \;\; -eq(2)$

• Finally, the % change in weight will be given by,

$\sf :\implies \delta W = \delta m + \delta g$

$\sf :\implies \delta W = \dfrac{100h}{R}$

$\displaystyle \underline{\bigstar\:\textsf{According to the Question :}}$

$\sf\dashrightarrow \delta W = \dfrac{100\times 32}{6400}$

$\dashrightarrow\underline{\boxed{\pink{\mathfrak {\delta w = 0.5 \%}}}}$

Answer 2

$\large{\rm{Aɴsᴡᴇʀ :}}$

$\:$

$\rm{\:\:•\:Percentage\: decrease\: in \:weight =}$$\rm{\textbf{\red{0.5\: \%}}}$

$\:$

$\large{\rm{Gɪᴠᴇɴ :}}$

$\:$

$\sf{\:\:•\: Depth\:(d) = 32\:km}$

$\sf{\:\:•\: Radius \:of\:earth\:(R) = 6400\:km}$

$\:$

$\large{\rm{Tᴏ \:Fɪɴᴅ :}}$

$\:$

$\sf{\:\:•\: Percentage\: decrease \: in \: weight \:of \:the \:body.}$

$\:$

$\large{\rm{Fᴏʀᴍᴜʟᴀ\: Usᴇᴅ :}}$

$\:$

$\:\:•\:$$\boxed{\sf{g' (g\:at\:depth\:‘d’) = g\left(1-\dfrac{d}{R}\right)}}$

$\:$

$\large{\rm{Sᴏʟᴜᴛɪᴏɴ :}}$

$\:$

➪ $\rm{Weight\:at\:depth = mg'}$

$\rm{ = mg\left(1-\dfrac{d}{R}\right)}$

$\:$

$\rm{➪\:Percentage\:decrease\:in\:weight = \dfrac{mg-mg'}{mg}\times 100}$

$\rm{ = \dfrac{mg-mg\left(1-\dfrac{d}{R}\right)}{mg}\times 100}$

$\rm{ = \left[\cancel{1}-\left(\cancel{1}-\dfrac{d}{R}\right)\right]\times 100}$

$\rm{ = \left(\dfrac{32}{64\cancel{00}}\right)\times 1\cancel{00}\: = }$$\rm{\red{ 0.5\: \%}}$